Question

If A=-1 -2 -2
2 1 -2
2 -2 1
then show that the adjoint of A is 3A'. Find A-1

​

Answer 1

We have:

$A=\left[\begin{array}{ccc}-1&-2&-2\\2&1&-2\\2&-2&1\end{array}\right]$

We have to show that adjoint of A = 3A'.

Also, find $A^{-1}$.

Solution:

∴ $A=\left[\begin{array}{ccc}-1&-2&-2\\2&1&-2\\2&-2&1\end{array}\right]$

[A] = - 1(1 - 4) + 2(2 + 4) - 2( - 4 - 2)

    = 3 + 12 + 12 = 27 ≠ 0, $A^{'}$ exists.

Adjoint of $A=\left[\begin{array}{ccc}-3&-6&-6\\6&3&-6\\-6&-6&3\end{array}\right]^'$

∴ Adjoint of $A=\left[\begin{array}{ccc}-3&6&6\\-6&3&-6\\-6&-6&3\end{array}\right]$

$A^{'} =\left[\begin{array}{ccc}-1&2&2\\-2&1&-2\\-2&-2&1\end{array}\right]$

∴ $3A^{'} =\left[\begin{array}{ccc}-3&6&6\\-6&3&-6\\-6&-6&3\end{array}\right]$

Thus, adjoint of A = 3A' = $\left[\begin{array}{ccc}-3&6&6\\-6&3&-6\\-6&-6&3\end{array}\right]$, shown.

We know that:

$A^{-1} =\dfrac{1}{[A]} (Adj A)$

∴ $A^{-1} =\dfrac{1}{9} \left[\begin{array}{ccc}-1&-2&-2\\2&1&-2\\2&-2&1\end{array}\right]$

$A^{-1} =\dfrac{1}{9} \left[\begin{array}{ccc}\dfrac{-1}{9}&\dfrac{-2}{9}&\dfrac{-2}{9}\\\dfrac{2}{9}&\dfrac{1}{9}&\dfrac{-2}{9}\\\dfrac{2}{9}&\dfrac{-2}{9}&\dfrac{1}{9}\end{array}\right]$

∴   $A^{-1} =\dfrac{1}{9} \left[\begin{array}{ccc}\dfrac{-1}{9}&\dfrac{-2}{9}&\dfrac{-2}{9}\\\dfrac{2}{9}&\dfrac{1}{9}&\dfrac{-2}{9}\\\dfrac{2}{9}&\dfrac{-2}{9}&\dfrac{1}{9}\end{array}\right]$

Answer 2

Given :  $A=\left[\begin{array}{ccc}-1&-2&-2\\2&1&-2\\2&-2&1\end{array}\right]$

To Find : Show that Adj(A) = 3A'

Find A⁻¹

Solution:

$A=\left[\begin{array}{ccc}-1&-2&-2\\2&1&-2\\2&-2&1\end{array}\right]$

A' is  Transpose of A

$A'=\left[\begin{array}{ccc}-1&2&2\\-2&1&-2\\-2&-2&1\end{array}\right]$

$A=\left[\begin{array}{ccc}-1&-2&-2\\2&1&-2\\2&-2&1\end{array}\right]$

| A |  =  -1 (1 - 4) + 2( 2 + 4)  - 2 ( - 4 - 2)

= 3 + 12 + 12

= 27

$A=\left[\begin{array}{ccc}-1&-2&-2\\2&1&-2\\2&-2&1\end{array}\right]$

$Adj A=\left[\begin{array}{ccc}C_{11}&C_{21}&C_{31}\\C_{12}&C_{22}&C_{32}\\C_{13}&C_{231}&C_{33}\end{array}\right]$

C₁₁ = (-1)¹⁺¹ (1*1 - (-2)(-2)) = - 3

C₁₂ = (-1)¹⁺² (2*1 - (2)(-2)) = - 6

C₁₃ = (-1)¹⁺³ (2*(-2) - (2)(1)) = - 6

and so on..

$Adj A=\left[\begin{array}{ccc}-3&6&6\\-6&3&-6\\-6&-6&3\end{array}\right]$

$Adj A=3\left[\begin{array}{ccc}-1&2&2\\-2&1&-2\\-2&-2&1\end{array}\right]$

Adj A = 3A'

A⁻¹  =( 1/|A| )Adj A  =( 1/27 )3A'

=> A⁻¹  = ( 1/9 )A'

$A^{-1}=\dfrac{1}{9}\left[\begin{array}{ccc}-1&2&2\\-2&1&-2\\-2&-2&1\end{array}\right]$

$A^{-1}=\left[\begin{array}{ccc}-1/9&2/9&2/9\\-2/9&1/9&-2/9\\-2/9&-2/9&1/9\end{array}\right]$

Learn More:

Find the inverse of the following matrix by using row elementary ...

https://brainly.in/question/18338344

If a=matrix 2,3,5,-2 write a inverse in terms of a - Brainly.in

https://brainly.in/question/3109606